技術標籤：coding面經

摘自889題解：https://leetcode-cn.com/problems/construct-binary-tree-from-preorder-and-postorder-traversal/solution/kan-wo-jiu-gou-liao-san-chong-bian-li-fang-shi-gou/

給前序+後序重構二叉樹：

LC889

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
 

class Solution {
public:
    TreeNode* fun(vector<int>& pre, vector<int>& post,int preBegin,int preEnd, int postBegin, int postEnd){
       if(preBegin > preEnd || postBegin > postEnd) return NULL;
       TreeNode *root = new TreeNode(pre[preBegin]);
       if(preBegin == preEnd )
 
 return root;
       int index = postBegin;
       while(post[index] != pre[preBegin+1]){
           index++;
       }
       root->left = fun(pre,post,preBegin+1,preBegin+1+(index-postBegin),postBegin,index);
       root->right = fun(pre,post,preBegin+1+(index-postBegin)+1,preEnd,index+1,postEnd-
 
1);
       return root;
    }

    TreeNode* constructFromPrePost(vector<int>& pre, vector<int>& post) {
        return fun(pre,post,0,pre.size()-1,0,post.size()-1);

    }
};

給前序+中序重構二叉樹：

LC105
注意：
1. 事實證明邊界條件必須是大於號：if(preBegin > preEnd || inBegin > inEnd)
2. 注意遞迴的邊界這裡要index-1-inbegin,上一個給前序後序的也不一樣，本質都是左子樹根位置-左子樹起始位置。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* fun(vector<int> &pre, vector<int> &in, int preBegin,int preEnd,int inBegin,int inEnd){
        if(preBegin > preEnd || inBegin > inEnd) return nullptr;
        TreeNode *root = new TreeNode(pre[preBegin]);
        if(preBegin == preEnd) return root;
        int index = inBegin;
        while(in[index] != pre[preBegin]){
            index++;
        }
        root->left = fun(pre,in,preBegin+1,preBegin+1+(index-1-inBegin), inBegin,index-1);
        root->right = fun(pre,in,preBegin+1+(index-1-inBegin)+1,preEnd,index+1,inEnd);
        return root;
    }
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        return fun(preorder,inorder,0,preorder.size()-1,0,inorder.size()-1);
    }
};

給中序+後序重建二叉樹：

LC106

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* fun(vector<int> &in,vector<int> &post,int inBegin,int inEnd,int postBegin,int postEnd){
        if(inBegin>inEnd || postBegin > postEnd) return NULL;
        TreeNode *root = new TreeNode(post[postEnd]);
        if(postBegin == postEnd) return root;
        int index = inBegin;
        while(in[index] != post[postEnd]){
            index++;
        }
        root->left = fun(in,post,inBegin,index-1,postBegin,postBegin+(index-1-inBegin));
        root->right = fun(in,post,index+1,inEnd,postBegin+(index-1-inBegin)+1,postEnd-1);
        return root;
    }
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        return fun(inorder,postorder,0,inorder.size()-1,0,postorder.size()-1);
    }
};